Given the following diagram:

enter image description here


  • W = (-1, 0)
  • X = (-1, 2)
  • Y = (1, 2)
  • Z = (1, 0)

How can I find M?

The ellipse can be assumed to be a semi-ellipse with one of the foci on $\bar{XY}$. I'm guessing that this means that one focus is at (0, 2), with the other focus at (0, -2). Now, by the definition of an ellipse, I know that the sum of the distances from those two points to any other point on the ellipse is a constant. But, having reasoned that far, I've hit a dead end. Where do I go from here?


Recall the equation of an ellipse: $\boxed{\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1}$

Furthermore, recall that if $c$ is the distance from the focus to the vertex, then: $\boxed{c^2=a^2-b^2}$

Since the focus is on $XY$, we know that $c=2$ so that $2^2=a^2-b^2$ or $\boxed{b^2=a^2-4}$.

Since $Y(1,2)$ is on the ellipse, we may substitute it into its equation, which yields: $$ \begin{align*} \dfrac{1^2}{a^2} + \dfrac{2^2}{b^2} &= 1 \\ \dfrac{1}{a^2} + \dfrac{4}{a^2-4} &= 1 \\ 1(a^2-4)+4(a^2) &=a^2(a^2-4) \\ 5a^2-4 &=a^4-4a^2 \\ 0 &=a^4-9a^2+4 \\ a^2 &= \dfrac{9 \pm \sqrt{65}}{2} \\ a &= \sqrt{\dfrac{9 \pm \sqrt{65}}{2}} \quad (\approx 0.6847 \text{ or } 2.9208) \end{align*} $$

However, notice that from the diagram, $M(a,0)$ lies to the right of $Z(1,0)$, implying that $a>1$. Hence, we reject the negative version and conclude that: $$ \boxed{a = \sqrt{\dfrac{9 + \sqrt{65}}{2}} = 2.9208...} $$

  • $\begingroup$ The OP said "I'm guessing that this means that one focus is at (0, 2)". In other words, this is not known to be true. If it is, your solution is correct. $\endgroup$ – marty cohen May 28 '13 at 0:10
  • $\begingroup$ Yes, the focus is at (0, 2), since the focus has to be on the midpoint of $\bar{XY}$. I wasn't sure of that when I posted the question, but I re-read the problem and I'm sure of that now. Also, thanks Adriano for your clear explanation. I'm not sure why I didn't think to substitute in the coordinates of a point, but you made it look obvious! $\endgroup$ – quanticle May 29 '13 at 5:23

The equation of an ellipse in standard form is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.

If it passes through $(1, 2)$, then $\frac{1}{a^2} + \frac{4}{b^2} = 1$.

However, this is one equation in two unknowns ($a$ and $b$). To determine the unknowns, you need another condition.

To see this visually, imagine M getting closer to Z. Then the upper vertex of the ellipse will move higher, and similarly as M moves further from Z.


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